$40-\:-8+4-3$
$3x+21<25$
$\sqrt{3x^2}.4\sqrt{10x^2}$
$\frac{dy}{dx}=\frac{-y^2}{1+x}$
$=\frac{1}{x^4}-\frac{3}{x^2}+x^3$
$5x^2-2x>8+4x^2$
$\left|\sqrt{36}\right|-\left|-20\right|-\left|3x-7\right|$
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